Lemma 10.136.18. Let $R$ be a ring and let $I \subset R$ be an ideal. Let $R/I \to \overline{S}$ be a syntomic map. Then there exists elements $\overline{g}_ i \in \overline{S}$ which generate the unit ideal of $\overline{S}$ such that each $\overline{S}_{g_ i} \cong S_ i/IS_ i$ for some relative global complete intersection $S_ i$ over $R$.

Proof. By Lemma 10.136.15 we find a collection of elements $\overline{g}_ i \in \overline{S}$ which generate the unit ideal of $\overline{S}$ such that each $\overline{S}_{g_ i}$ is a relative global complete intersection over $R/I$. Hence we may assume that $\overline{S}$ is a relative global complete intersection. Write $\overline{S} = (R/I)[x_1, \ldots , x_ n]/(\overline{f}_1, \ldots , \overline{f}_ c)$ as in Definition 10.136.5. Choose $f_1, \ldots , f_ c \in R[x_1, \ldots , x_ n]$ lifting $\overline{f}_1, \ldots , \overline{f}_ c$. Set $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$. Note that $S/IS \cong \overline{S}$. By Lemma 10.136.11 we can find $g \in S$ mapping to $1$ in $\overline{S}$ such that $S_ g$ is a relative global complete intersection over $R$. Since $\overline{S} \cong S_ g/IS_ g$ this finishes the proof. $\square$

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